Integrand size = 45, antiderivative size = 653 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {(a-b) \sqrt {a+b} \left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{24 a b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (24 A b^2-18 a b B+12 b^2 B+15 a^2 C-10 a b C+16 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{24 b^3 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (6 a^2 b B+8 b^3 B-5 a^3 C-4 a b^2 (2 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{8 b^4 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 b^3 d} \]
1/3*C*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(3/2)+1/12*(6*B*b-5 *C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d/sec(d*x+c)^(1/2)+1/24*(24*A* b^2-18*B*a*b+15*C*a^2+16*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+ c)^(1/2)/b^3/d-1/24*(a-b)*(24*A*b^2-18*B*a*b+15*C*a^2+16*C*b^2)*csc(d*x+c) *EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a- b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a* (1+sec(d*x+c))/(a-b))^(1/2)/a/b^3/d/sec(d*x+c)^(1/2)+1/24*(24*A*b^2-18*B*a *b+12*B*b^2+15*C*a^2-10*C*a*b+16*C*b^2)*csc(d*x+c)*EllipticF((a+b*cos(d*x+ c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*c os(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1 /2)/b^3/d/sec(d*x+c)^(1/2)-1/8*(6*B*a^2*b+8*B*b^3-5*a^3*C-4*a*b^2*(2*A+C)) *csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2) ,(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+ c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1818\) vs. \(2(653)=1306\).
Time = 20.05 (sec) , antiderivative size = 1818, normalized size of antiderivative = 2.78 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[a + b*Cos[c + d*x] ]*Sec[c + d*x]^(3/2)),x]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sin[c + d*x])/(12*b) + (( 6*b*B - 5*a*C)*Sin[2*(c + d*x)])/(24*b^2) + (C*Sin[3*(c + d*x)])/(12*b)))/ d + (Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^ 2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-24*a*A*b^2*Tan[(c + d*x)/2] - 24*A*b^3*Tan[(c + d*x)/2] + 18*a^2*b*B*Tan[(c + d*x)/2] + 18*a*b ^2*B*Tan[(c + d*x)/2] - 15*a^3*C*Tan[(c + d*x)/2] - 15*a^2*b*C*Tan[(c + d* x)/2] - 16*a*b^2*C*Tan[(c + d*x)/2] - 16*b^3*C*Tan[(c + d*x)/2] + 48*A*b^3 *Tan[(c + d*x)/2]^3 - 36*a*b^2*B*Tan[(c + d*x)/2]^3 + 30*a^2*b*C*Tan[(c + d*x)/2]^3 + 32*b^3*C*Tan[(c + d*x)/2]^3 + 24*a*A*b^2*Tan[(c + d*x)/2]^5 - 24*A*b^3*Tan[(c + d*x)/2]^5 - 18*a^2*b*B*Tan[(c + d*x)/2]^5 + 18*a*b^2*B*T an[(c + d*x)/2]^5 + 15*a^3*C*Tan[(c + d*x)/2]^5 - 15*a^2*b*C*Tan[(c + d*x) /2]^5 + 16*a*b^2*C*Tan[(c + d*x)/2]^5 - 16*b^3*C*Tan[(c + d*x)/2]^5 + 48*a *A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/ 2]^2)/(a + b)] - 36*a^2*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2] ^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 48*b^3*B*EllipticPi[-1, ArcSin[Tan[( c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*a^3*C*Elliptic Pi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*...
Time = 3.10 (sec) , antiderivative size = 623, normalized size of antiderivative = 0.95, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4709, 3042, 3528, 27, 3042, 3528, 27, 3042, 3540, 25, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sec (c+d x)^{3/2} \sqrt {a+b \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\sqrt {a+b \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} \left ((6 b B-5 a C) \cos ^2(c+d x)+2 b (3 A+2 C) \cos (c+d x)+3 a C\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} \left ((6 b B-5 a C) \cos ^2(c+d x)+2 b (3 A+2 C) \cos (c+d x)+3 a C\right )}{\sqrt {a+b \cos (c+d x)}}dx}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left ((6 b B-5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (3 A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a C\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right ) \cos ^2(c+d x)+2 b (6 b B+a C) \cos (c+d x)+a (6 b B-5 a C)}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right ) \cos ^2(c+d x)+2 b (6 b B+a C) \cos (c+d x)+a (6 b B-5 a C)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (6 b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )+a (6 b B-5 a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3540 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int -\frac {-3 \left (-5 C a^3+6 b B a^2-4 b^2 (2 A+C) a+8 b^3 B\right ) \cos ^2(c+d x)-2 a b (6 b B-5 a C) \cos (c+d x)+a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-3 \left (-5 C a^3+6 b B a^2-4 b^2 (2 A+C) a+8 b^3 B\right ) \cos ^2(c+d x)-2 a b (6 b B-5 a C) \cos (c+d x)+a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-3 \left (-5 C a^3+6 b B a^2-4 b^2 (2 A+C) a+8 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (6 b B-5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )-2 a b (6 b B-5 a C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-3 \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )-2 a b (6 b B-5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a \left (15 C a^2-18 b B a+24 A b^2+16 b^2 C\right )-2 a b (6 b B-5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sqrt {a+b} \cot (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {6 \sqrt {a+b} \cot (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {-a \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sqrt {a+b} \cot (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {6 \sqrt {a+b} \cot (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sin (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {-\frac {2 \sqrt {a+b} \cot (c+d x) \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}+\frac {6 \sqrt {a+b} \cot (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}}{4 b}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 b d}}{6 b}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Cos[c + d*x]^(3/2)*Sqrt[a + b*Co s[c + d*x]]*Sin[c + d*x])/(3*b*d) + (((6*b*B - 5*a*C)*Sqrt[Cos[c + d*x]]*S qrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*b*d) + (-1/2*((2*(a - b)*Sqrt[a + b]*(24*A*b^2 - 18*a*b*B + 15*a^2*C + 16*b^2*C)*Cot[c + d*x]*EllipticE[Arc Sin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/ (a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x])) /(a - b)])/(a*d) - (2*Sqrt[a + b]*(24*A*b^2 - 18*a*b*B + 12*b^2*B + 15*a^2 *C - 10*a*b*C + 16*b^2*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d + (6*Sqrt [a + b]*(6*a^2*b*B + 8*b^3*B - 5*a^3*C - 4*a*b^2*(2*A + C))*Cot[c + d*x]*E llipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos [c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqr t[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d))/b + ((24*A*b^2 - 18*a*b*B + 15*a ^2*C + 16*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(4*b))/(6*b))
3.16.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] & & NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(4745\) vs. \(2(593)=1186\).
Time = 13.61 (sec) , antiderivative size = 4746, normalized size of antiderivative = 7.27
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4746\) |
| default | \(\text {Expression too large to display}\) | \(4786\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(1/2 ),x,method=_RETURNVERBOSE)
A/d/b/(1+cos(d*x+c))/(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2)*(-(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ell ipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a-(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot (d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b+2*(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(cot(d*x+c) -csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) *(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc( d*x+c),(-(a-b)/(a+b))^(1/2))*a*sec(d*x+c)-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1 /2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-c sc(d*x+c),(-(a-b)/(a+b))^(1/2))*b*sec(d*x+c)+4*EllipticPi(cot(d*x+c)-csc(d *x+c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* (a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*sec(d*x+c)+b*sin(d*x+c)-(cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*E llipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*sec(d*x+c)^2-(cos(d *x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 )*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b*sec(d*x+c)^2+2*E llipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*sec(d*x +c)^2+a*tan(d*x+c))+1/4*B/d/b^2/(1+cos(d*x+c))/(a+b*cos(d*x+c))^(1/2)/s...
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(1/2),x, algorithm="fricas")
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/(sqrt(a + b*cos(c + d*x) )*sec(c + d*x)**(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(1/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c) + a )*sec(d*x + c)^(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(1/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c) + a )*sec(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + b *cos(c + d*x))^(1/2)),x)